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\title{Chapter 11: CHARACTERISTIC VARIETIES}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

\begin{document}

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% 封面页
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  \titlepage
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% 目录页
\begin{frame}{Contents}
  \tableofcontents
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% Section 0
%\section{INTRO.}
%---------------------------------------------------
\begin{frame}{intro. }

In this chapter we give a geometrical interpretation of the dimension of an $A_n$-module. 

In order to do this we will have to use a number of results of algebraic geometry and linear symplectic geometry. 

All the algebraic geometry that we need can be found in [Hartshorne 77, Ch. 1]. 

Throughout the chapter we assume that the base field is $\mathbb{C}$.

\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 1
\section{THE CHARACTERISTIC VARIETY.}
%---------------------------------------------------
\begin{frame}[allowframebreaks]{A. }

Let $A_n$ be the $n$-th complex Weyl algebra and let $M$ be a finitely generated left $A_n$-module with a good filtration $\Gamma$. 

Then $\text{gr}^{\Gamma}M$ is a finitely generated module over the polynomial ring $S_n$. 

Let $\text{ann}(M,\Gamma)$ stand for the annihilator of $\text{gr}^{\Gamma}M$ in $S_n$. 

Then $\text{ann}(M,\Gamma)$ is an ideal of $S_n$. 

It depends not only on $M$, but also on the choice of the good filtration $\Gamma$; see Exercise 4.1. 

The radical of $\text{ann}(M,\Gamma)$ however is independent of the filtration.

\textbf{1.1. Lemma.}
Let $\Omega$ be another good filtration of $M$. Then
$$
\text{rad}(\text{ann}(M,\Gamma)) = \text{rad}(\text{ann}(M,\Omega)).
$$
%\end{lemma}

\textbf{Proof.} 
The ideals $\text{ann}(M,\Gamma)$ and $\text{ann}(M,\Omega)$ are homogeneous (see Exercise 8.4.6), hence so are their radicals (Exercise 4.2). 

Choose a homogeneous element $f$ of degree $s$ in $\text{rad}(\text{ann}(M,\Gamma))$. 

There exists $d \in B_s$, the component of degree $\leq s$ of the Bernstein filtration, such that $f = \sigma_s(d)$.

Since $f \in \text{rad}(\text{ann}(M,\Gamma))$, there exists $m \in \mathbb{N}$ such that $f^m \in \text{ann}(M,\Gamma)$. 

Hence, $d^m\Gamma_i \subseteq \Gamma_{ms+i-1}$, for every $i \geq 0$. 

Iterating $q$ times we get that
\begin{equation}
d^{mq}\Gamma_i \subseteq \Gamma_{i+msq-q}.
\tag{1.2}
\end{equation}

On the other hand, by Proposition 8.3.2 there exists $k \geq 0$ such that
$$
\Gamma_{i-k} \subseteq \Omega_i \subseteq \Gamma_{i+k},
$$
for all $i \geq 0$. 

Together with (1.2) for $q = 2k + 1$, this leads to
$$
d^{m(2k+1)}\Omega_i \subseteq d^{m(2k+1)}\Gamma_{i+k} \subseteq \Gamma_{i+ms(2k+1)-k-1} \subseteq \Omega_{i+ms(2k+1)-1}.
$$

Thus $d^{m(2k+1)}\Omega_i \subseteq \Omega_{i+ms(2k+1)-1}$, for all $i \geq 0$. 

Hence $f^{m(2k+1)} \in \text{ann}(M,\Omega)$, and so $\text{rad}(\text{ann}(M,\Gamma)) \subseteq \text{rad}(\text{ann}(M,\Omega))$. 

The opposite inclusion follows by swapping $\Gamma$ and $\Omega$.


The ideal $\mathcal{I}(M) = \text{rad}(\text{ann}(M,\Gamma))$ is called the {\color{red} characteristic ideal} of $M$. 

It follows from Lemma 1.1 that it is independent of the good filtration $\Gamma$ used to calculate it. 

In other words, $\mathcal{I}(M)$ is an invariant of $M$. 

Thus, so is the affine variety
$$
\text{Ch}(M) = \mathcal{Z}(\mathcal{I}(M)) \subseteq \mathbb{C}^{2n}
$$
that it determines. 

This variety is called the {\color{red} characteristic variety} of $M$. 

Since $\mathcal{I}(M)$ is a homogeneous ideal (see Exercise 4.2), the variety $\text{Ch}(M)$ is homogeneous. 

This means that if $p \in \text{Ch}(M)$ and $\lambda \in \mathbb{C}$, then $\lambda p \in \text{Ch}(M)$. 

Note that $\text{Ch}(M)$ is a subvariety of $\mathbb{C}^{2n}$, since $S_n$ is a polynomial ring in $2n$ variables!


Here is a simple example. 

Let $d \in A_n$ be an element of degree $r$ and put $M = A_n/A_nd$. 

If $\mathcal{B}''$ is the filtration of $M$ induced by the Bernstein filtration of $A_n$, then
$$
\text{gr}^{\mathcal{B}''}M = S_n/S_n\sigma_r(d)
$$
as we saw in Ch. 7, §5. 

Therefore, $\text{ann}(M,\mathcal{B}'') = S_n\sigma_r(d)$ and so $\text{Ch}(M) = \mathcal{Z}(\sigma_r(d))$ is a hypersurface.


\textbf{1.3. Proposition. }
Let $M$ be a finitely generated left $A_n$-module and $N$ a submodule of $M$. Then
$$
\text{Ch}(M) = \text{Ch}(N) \cup \text{Ch}(M/N).
$$
%\end{proposition}


\textbf{Proof.} 
Let $\Gamma$ be a good filtration of $M$. 

It induces good filtrations $\Gamma'$ and $\Gamma''$ in $N$ and $M/N$ respectively. 

By Lemma 7.5.1, there exists an exact sequence of finitely generated graded $S_n$-modules,
$$
0 \to \text{gr}^{\Gamma''}N \to \text{gr}^{\Gamma'}M \to \text{gr}^{\Gamma''}M/N \to 0.
$$

Clearly
$$
\text{ann}(M,\Gamma) \subseteq \text{ann}(N,\Gamma') \cap \text{ann}(M/N,\Gamma'')
$$
and so $\text{Ch}(N) \cup \text{Ch}(M/N) \subseteq \text{Ch}(M)$, by [Hartshorne 77, Ch. 1, Proposition 1.2]. 

The opposite inclusion follows from
$$
\text{ann}(N,\Gamma) \text{ann}(M/N,\Gamma'') \subseteq \text{ann}(M,\Gamma).
$$


Before we proceed with the study of the characteristic variety, let us review some basic facts about the dimension of an affine variety. 

Let $J$ be an ideal of $S_n = \mathbb{C}[y_1,\ldots,y_{2n}]$ and put $V = \mathcal{Z}(J)$. 

If $p$ is a point of $V$ then the Zariski tangent space of $V$ at $p$ is the linear subspace of $\mathbb{C}^{2n}$ defined by the equations
$$
\sum_{1}^{2n} \frac{\partial F}{\partial y_i}(p)y_i = 0
$$
for all $F \in J$. 

This space is denoted by $T_p(V)$; it is a complex vector subspace of $\mathbb{C}^{2n}$.


The dimension of $V$ equals $\inf\{\dim_{\mathbb{C}} T_p(V) : p \in V\}$. 

Actually one need not look at every point of $V$. 

If $p$ is a non-singular point of $V$, then $\dim(V) = \dim_{\mathbb{C}} T_p(V)$. 

This is equivalent to the definition in terms of heights of prime ideals, see [Hartshorne 77, Ch. 1, Exercise 5.10].


We may also define $T_p(V)$ in terms of {\color{red}linear forms} on $\mathbb{C}^{2n}$. 

Let $d_pf$ be the {\color{red}linear form} defined on the vector $Y = (y_1,\ldots,y_{2n})$ by
$$
d_pf(Y) = \sum_{1}^{2n} \frac{\partial f}{\partial y_i}(p)y_i.
$$

Consider the map $d_p: S_n \to (\mathbb{C}^{2n})^*$ defined by $d_p(f) = d_pf$. 

Let $f,g \in S_n$ and $\lambda \in \mathbb{C}$. 

Then $d_p$ satisfies
\begin{align*}
d_p(f + \lambda g) &= d_pf + \lambda d_pg, \\
d_p(fg) &= f(p)d_pg + g(p)d_pf.
\end{align*}

We may rephrase the definition of $T_p(V)$ using $d_p$, as follows: 
$u \in T_p(V)$ if and only if $d_pF(u) = 0$, for every $F \in \mathcal{I}(V)$. 

This may be improved, by restricting the elements of $\mathcal{I}(V)$ to a finite set. 

Suppose that $F_1,\ldots,F_m$ generate $\mathcal{I}(V)$. 

It follows from the properties of $d_p$ stated above that $u \in T_p(V)$ if and only if
$$
d_pF_1(u) = \cdots = d_pF_m(u) = 0.
$$

The characterization of $T_p(V)$ in terms of $d_p$ will be used in the next section.


The following theorem is an immediate consequence of the fact that if $N$ is a graded module over $S_n$ 
then the degree of its Hilbert polynomial is $\dim \mathcal{Z}(\text{ann}_{S_n} N)$; see [Hartshorne 77, Ch. 1 Theorem 7.5].


\textbf{1.4. Theorem.}
Let $M$ be a finitely generated left module over $A_n$. Then $\dim \text{Ch}(M) = d(M)$.
%\end{theorem}


It is now very easy to show that if $d \neq 0$ is an operator in $A_n$, then $A_n/A_nd$ has dimension $2n-1$. 

Suppose that $d$ has degree $r > 0$. 

As we have seen, the characteristic variety of $A_n/A_nd$ has equation $\sigma_r(d) = 0$. 

This is a hypersurface of $\mathbb{C}^{2n}$; hence its dimension is $2n-1$. 

Thus $d(A_n/A_nd) = 2n-1$, by Theorem 1.4. 

Compare this example with Exercise 9.5.2.


We may also use the characteristic variety to achieve a geometrical interpretation of Bernstein's inequality. 

This depends on results of symplectic geometry that we summarize in the next section.

\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 2
\section{SYMPLECTIC GEOMETRY.}
%---------------------------------------------------
\begin{frame}[allowframebreaks]{B. }

A symplectic structure on $\mathbb{C}^{2n}$ is determined by a non-degenerate skew-symmetric form. 

The standard symplectic structure on $\mathbb{C}^{2n}$ is expressed by means of the matrix
$$
\Omega = \begin{pmatrix}
0 & -\mathrm{I}_n \\
\mathrm{I}_n & 0
\end{pmatrix}
$$
where $\mathrm{I}_n$ is the $n \times n$ identity matrix. 

Given vectors $u,v \in \mathbb{C}^{2n}$, the {\color{red}symplectic form} is
$$
\omega(u,v) = u\Omega v^t.
$$

The matrix of any {\color{red}non-degenerate skew-symmetric form} can be written in the form above for a suitable choice of coordinates [Cohn 84, §8.6, Theorem 1].

If $U$ is a subspace of $\mathbb{C}^{2n}$, then its {\color{red}skew-orthogonal complement} is
$$
U^\perp = \{v \in \mathbb{C}^{2n} : \omega(u,v) = 0 \text{ for all } u \in U\}.
$$

Note that it is not true that $\mathbb{C}^{2n}$ is the direct sum of a subspace and its skew-orthogonal complement. 

For example, since $\omega(u,u) = 0$ for every $u \in \mathbb{C}^{2n}$, then every one-dimensional space is contained in its complement. 

A subspace that is contained in its skew-symmetric complement is called {\color{red} isotropic}. 

However, since $\omega$ is non-degenerate, it is always true that $U$ and $U^\perp$ have complementary dimensions. 

{\color{blue}My proof. Let $u_1,\cdots,u_k$ be a basis of $U$, let $A=(u_1^t,\cdots,u_k^t)$ be the $2n\times k$ matrix. Then $U^\perp$ is the solution space of $v$ of the system of linear equations $v\Omega A=0$. Since the rank of the matrix $\Omega A$ is $k$, we see $U^\perp$ has dimension $2n-k$. }

The proof is short enough to be included here. First, some definitions. 

For $w \in \mathbb{C}^{2n}$, define a linear form $\phi_w: \mathbb{C}^{2n} \to \mathbb{C}$ by $\phi_w(v) = \omega(w,v)$. 

Let $\Phi: \mathbb{C}^{2n} \to (\mathbb{C}^{2n})^*$ be the linear map which associates $\phi_w$ to $w \in \mathbb{C}^{2n}$.

\textbf{2.1. Proposition.}
Let $\omega$ be a non-degenerate skew-symmetric form in $\mathbb{C}^{2n}$ and $U$ a subspace of $\mathbb{C}^{2n}$. Then:
\begin{enumerate}
    \item $\Phi$ is an isomorphism.
    \item If a linear form $\phi_w$ restricts to zero on $U$, then $w \in U^\perp$.
    \item $\dim U + \dim U^\perp = 2n$.
\end{enumerate}
%\end{proposition}

\textbf{Proof.} Note that the kernel of $\Phi$ is zero, because $\omega$ is non-degenerate. 

Hence,
$$
\dim \Phi(\mathbb{C}^{2n}) = 2n - \dim \ker(\Phi) = 2n.
$$

Since $\dim (\mathbb{C}^{2n})^* = 2n$, it follows that $\Phi$ is surjective. 

Thus $\Phi$ is an isomorphism of vector spaces, and (1) is proved.

Now let $\Phi|_U: \mathbb{C}^{2n} \to U^*$ be the map defined by $\Phi|_U(w) = \phi_w|_U$, the restriction of $\phi_w$ to $U$. 

Since every linear form on $U$ extends to a linear form on $\mathbb{C}^{2n}$, we have that $\Phi|_U$ is surjective by (1). 

Hence the image of $\Phi|_U$ is $U^*$.

Now (2) is clear, and it implies that the kernel of $\Phi|_U$ is $U^\perp$. 

Hence,
$$
2n = \dim \ker \Phi|_U + \dim \Phi|_U(\mathbb{C}^{2n}) = \dim U^\perp + \dim U^*,
$$
which implies (3).


The subspaces that are important, from our point of view, are the ones that contain their skew-complement. 

They are called {\color{red} co-isotropic} or {\color{red} involutive}. 

A hyperplane is always an {\color{red} involutive} subspace. 

Indeed, if $H$ is a hyperplane of $\mathbb{C}^{2n}$ that is not {\color{red} involutive}, then there exists $v \notin H$ such that $\omega(v,u) = 0$, for all $u \in H$. 

Since $\mathbb{C}^{2n} = H + \mathbb{C}v$, this contradicts the non-degeneracy of $\omega$.


If $V$ is an affine variety of $\mathbb{C}^{2n}$, then we will say that it is {\color{red} involutive} 
if the tangent space $T_p(V) \subseteq \mathbb{C}^{2n}$ is {\color{red} involutive} 
at every non-singular point $p \in V$.

In particular, a hypersurface will always be {\color{red} involutive}. 

The dimension of an {\color{red} involutive} variety satisfies an inequality similar to Bernstein's.


\textbf{2.2. Proposition.}
The dimension of an {\color{red} involutive} variety of $\mathbb{C}^{2n}$ is greater than or equal to $n$.
%\end{proposition}

\textbf{Proof.} Let $V$ be an {\color{red} involutive} variety of $\mathbb{C}^{2n}$ and $p$ a non-singular point of $V$. 

By definition, we have that $T_p(V)^\perp \subseteq T_p(V)$. 

Hence $\dim T_p(V)^\perp \leq \dim T_p(V)$. 

Thus, by Proposition 2.1,
$$
2n = \dim T_p(V)^\perp + \dim T_p(V) \leq 2 \dim T_p(V),
$$
and so $\dim T_p(V) \geq n$. 

Therefore, $\dim(V) \geq n$.


The {\color{red} involutive} varieties of dimension $n$ are called {\color{red} lagrangian}. 

Note that if $V$ is {\color{red}lagrangian} in $\mathbb{C}^{2n}$, then $T_p(V)^\perp \subseteq T_p(V)$ and both subspaces have dimension $n$. 

Hence $T_p(V) = T_p(V)^\perp$. 

In particular, $V$ is also {\color{red}isotropic}. 

Thus {\color{red}lagrangian varieties} can also be characterized as varieties which are {\color{red} involutive} (co-isotropic) and {\color{red}isotropic}.


Talk of an affine variety, and one immediately thinks of its defining ideal. 

How can one decide whether a variety is {\color{red} involutive} by looking at its ideal? 

The answer lies with the Poisson bracket of $S_n$.


Let $I$ denote the inverse of the map $\Phi$ defined above, 
$$
\Phi: \mathbb{C}^{2n}  \to (\mathbb{C}^{2n})^*, \quad w  \mapsto \Phi(w)=\phi_w=\omega(w,-).
$$

The {\color{red} Poisson bracket} of $f,g \in S_n$ is
$$
\{f,g\}(p) = \omega(Id_pf,Id_pg) = d_pf(Id_pg).
$$

An explicit calculation using coordinates shows that
$$
\{f,g\}(p) = \sum_{i=1}^{n}\left\{\frac{\partial f}{\partial y_{n+i}}(p) \cdot \frac{\partial g}{\partial y_i}(p) - \frac{\partial g}{\partial y_{n+i}}(p) \cdot \frac{\partial f}{\partial y_i}(p)\right\}.
$$

From this formula it is easy to see that $\{f,g\}$ is a polynomial in $S_n$ and that the map $\{f,\cdot\}: S_n \to S_n$ is a derivation of $S_n$. 

An ideal $J$ of $S_n$ is {\color{red} closed} for the Poisson bracket if $\{f,g\} \in J$ whenever $f,g \in J$. 

We want to show that a variety is {\color{red} involutive} if its ideal is closed for the Poisson bracket. 

The proof will make use of a technical fact about the tangent space that we isolate in a lemma.


\textbf{2.3. Lemma.}
Let $V$ be an affine variety of $\mathbb{C}^{2n}$ and $p$ a point of $V$. 
If $\theta$ is a form on $\mathbb{C}^{2n}$ whose restriction to $T_p(V)$ is zero, 
then there exists $f \in \mathcal{I}(V)$ such that $\theta = d_pf$.
%\end{lemma}

\textbf{Proof.} Let $F_1,\ldots,F_m$ be the generators of $\mathcal{I}(V)$. 

Set $Y = (y_1,\ldots,y_{2n})$. 

Then $T_p(V)$ is the solution set of the equations
$$
d_pF_1(Y) = \cdots = d_pF_m(Y) = 0
$$
in $\mathbb{C}^{2n}$. 

But $\theta(Y) = 0$ is satisfied by the vectors of $T_p(V)$, by hypothesis. Thus
$$
\theta = \sum_{1}^{m} a_i d_pF_i,
$$
where $a_i \in \mathbb{C}$. Let $f = \sum_{1}^{m} a_i F_i \in \mathcal{I}(V)$. 

A straightforward calculation shows that $\theta = d_pf$.


\textbf{2.4. Proposition.}
An affine variety $V$ of $\mathbb{C}^{2n}$ is {\color{red} involutive} if and only if its ideal $\mathcal{I}(V)$ is closed for the Poisson bracket.
%\end{proposition}


\textbf{Proof.} Suppose that $V$ is {\color{red} involutive}. 

Let $p$ be a non-singular point of $V$. 

If $f \in \mathcal{I}(V)$, then $d_pf$ restricted to $T_p(V)$ is zero. 

By Proposition 2.1(2), $Id_pf \in T_p(V)^\perp$. 

Since $V$ is {\color{red} involutive}, then $Id_pf \in T_p(V)$. 

Now if $g \in \mathcal{I}(V)$, then $d_pg$ is zero on $T_p(V)$, hence
$$
\{f,g\}(p) = -d_pg(Id_pf) = 0.
$$

Since this identity holds for all non-singular points $p$ of $V$, we conclude that the polynomial $\{f,g\}$ is in $\mathcal{I}(V)$, which is then closed for the Poisson bracket.


Conversely, assume that $\mathcal{I}(V)$ is closed for the Poisson bracket. 

Choose $w \in T_p(V)^\perp$. 

By Lemma 2.3, there exists $g \in \mathcal{I}(V)$ such that $\phi_w = d_pg$. 

But $d_pg(u) = \omega(Id_pg,u)$. 

Since $\omega$ is non-degenerate, $w = Id_pg$. 

If $f \in \mathcal{I}(V)$ then
$$
0 = \{f,g\}(p) = d_pf(Id_pg) = d_pf(w).
$$

Hence $w \in T_p(V)$. 

Thus $T_p(V)^\perp \subseteq T_p(V)$.


We have seen that hypersurfaces are {\color{red} involutive}, since their tangent spaces are hyperplanes. 

This is very easy to prove using Proposition 2.4, since the ideal of a hypersurface is principal, and the Poisson bracket of a polynomial with itself is always zero.


The relation between characteristic varieties and symplectic geometry is the subject of the next theorem. 

The first proofs of this result were analytical. 

There is now a purely algebraic proof due to O. Gabber [Gabber 81]. 

We shall not include the proof here as it would take us too far from our intended course.


\textbf{2.5. Theorem.}
Let $M$ be a finitely generated left $A_n$-module. Then $\text{Ch}(M)$ is {\color{red} involutive} with respect to the standard symplectic structure of $\mathbb{C}^{2n}$. Equivalently, $\mathcal{I}(M)$ is closed for the Poisson bracket.
%\end{theorem}


It is not obvious at first sight why this theorem should be difficult to prove. 

In fact, it is easy to prove that if $\Gamma$ is a good filtration of $M$, then $\text{ann}(M,\Gamma)$ is closed for the Poisson bracket. 

The theorem does {\color{red} not} follow easily from this fact because it is {\color{red} not} true that the radical of a closed ideal is closed! 

Here is an example. 

Let $J$ be the ideal of $S_1$ generated by $y_1^2$, $y_2^2$ and $y_1y_2$. 

It is closed for the Poisson bracket, but its radical contains $y_1$ and $y_2$, and $\{y_1,y_2\}=1$. 

Thus $\text{rad}(J)$ is not closed for the Poisson bracket. 

Theorem 2.5 states the very subtle fact that the radical of the annihilator of the graded module of an $A_n$-module {\color{red} is} closed, even though radicals of closed ideals are {\color{red} not} in general closed.


Putting together Theorem 2.5 and Proposition 2.2 we get Bernstein's inequality.

\textbf{2.6. Corollary.}
Let $M$ be a finitely generated left $A_n$-module. Then
$$
d(M) = \dim \text{Ch}(M) \geq n.
$$
%\end{corollary}


It also follows from Theorem 2.5 that a holonomic module has a homogeneous lagrangian characteristic variety. 

A very spectacular use of the involutivity of the characteristic variety is discussed in the next section.


\end{frame}
%---------------------------------------------------

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 3
\section{NON-HOLONOMIC IRREDUCIBLE MODULES.}
%---------------------------------------------------
\begin{frame}[allowframebreaks]{C. }

It is a very curious fact that, until 1985, it was widely believed that every {\color{red}irreducible} $A_n$-module had to be holonomic. 

There was no good reason for this, except a lack of examples. 

The truth came to light in [Stafford 85]. 

Stafford showed that if $n > 2$ and $\lambda_2,\ldots,\lambda_n \in \mathbb{C}$ are algebraically independent over $\mathbb{Q}$, then the operator
$$
s = \partial_1 + \left(\sum_{2}^{n} \lambda_i x_i \partial_i + x_i\right) + \sum_{2}^{n}(x_i - \partial_i)
$$
generates a maximal left ideal of $A_n$. 

Let $M = A_n/A_ns$. 

Then $M$ is {\color{red}irreducible} and $d(M) = 2n-1 > n$, since $n > 2$. 

The fact that $A_ns$ is maximal is proved by a long calculation, which can also be found in [Krause and Lenagan 85, Theorem 8.7].


In 1988, J. Bernstein and V. Lunts found a different and more geometric way of constructing {\color{red}irreducible} modules of dimension $2n-1$ over $A_n$. 

As we have seen in §§1 and 2, the characteristic variety of an $A_n$-module is always homogeneous and {\color{red} involutive}. 

We will say that a homogeneous {\color{red} involutive} variety of $\mathbb{C}^{2n}$ is {\color{red} minimal} if it does not contain a proper {\color{red} involutive} homogeneous subvariety. 

For example, an {\color{red}irreducible} lagrangian variety must be {\color{red}minimal}, since varieties of dimension less than $n$ cannot be {\color{red} involutive}. 

The work of Bernstein and Lunts depends on the following result.


\textbf{3.1. Proposition.}
Let $d$ be an operator of degree $k$ in $A_n$ and suppose that:
\begin{enumerate}
    \item the symbol $\sigma_k(d)$ is {\color{red}irreducible} in $S_n$;
    \item the hypersurface $\mathcal{Z}(\sigma_k(d))$ is {\color{red}minimal}.
\end{enumerate}

Then the left ideal $A_nd$ is maximal. In particular, the quotient $A_n/A_nd$ is an {\color{red}irreducible} module of dimension $2n-1$ over $A_n$.
%\end{proposition}


\textbf{Proof.} Suppose that $J$ is a left ideal of $A_n$ such that $A_nd \subset J$. 

Since these are submodules of $A_n$ we may consider their graded modules with respect to the filtrations induced by $\mathcal{B}$; we get
$$
S_n \sigma_k(d) \subset \text{gr}(J) \subseteq S_n.
$$

The corresponding varieties are
$$
\emptyset \subseteq \mathcal{Z}(\text{gr}(J)) \subset \mathcal{Z}(\sigma_k(d)).
$$

Note that the last inclusion is proper, because $\sigma_k(d)$ is {\color{red}irreducible}. 

But $\text{gr}(A_n/J) \cong S_n/\text{gr}(J)$, and so $\text{ann}(A_n/J,\mathcal{B}) = \text{gr}(J)$. 

Therefore, $\mathcal{Z}(\text{gr}(J)) = \text{Ch}(A_n/J)$ is {\color{red} involutive} by Theorem 2.5. 

Since $\mathcal{Z}(\sigma_k(d))$ is {\color{red}minimal}, then $\mathcal{Z}(\text{gr}(J)) = \emptyset$. 

Hence $\text{gr}(J) = S_n$ and so $J = A_n$. 

Thus $A_nd$ is a maximal left ideal of $A_n$.


To put this proposition to good use we have only to construct {\color{red}minimal} hypersurfaces in $\mathbb{C}^{2n}$. 

That these hypersurfaces exist is the heart of the work of Bernstein and Lunts. 

In fact they show that most hypersurfaces in $\mathbb{C}^{2n}$ are {\color{red}minimal}. 

To make this into a precise statement we need a definition.


Let $S_n(k)$ be the homogeneous component of degree $k$ of $S_n$. 

This is a finite dimensional complex vector space; so it makes sense to talk about hypersurfaces in $S_n(k)$. 

We say that a property $\mathbf{P}$ holds {\color{red} generically} in $S_n(k)$ if the set
$$
\{f \in S_n(k) : \mathbf{P} \text{ does not hold for } f\}
$$
is contained in the union of countably many hypersurfaces of $S_n(k)$.


\textbf{3.2. Theorem.}
The property '$\mathcal{Z}(f)$ is {\color{red}minimal}' holds generically in $S_n(k)$, whenever $k \geq 4$ and $n \geq 2$.
%\end{theorem}


This result was proved for $n=2$ in [Bernstein and Lunts 88] and later generalized to $n \geq 2$ in [Lunts 89]. 

The proof of Theorem 3.2 uses a lot of algebraic geometry and also some results on differential equations, one of which goes back to Poincaré's thesis! 

The proficient (proficient means skilled and experienced) reader will find more details in the original papers.


The gist of the work of Bernstein and Lunts is that 'most' {\color{red}irreducible} $A_n$-modules are {\color{red} not} holonomic: almost the exact opposite of what was believed before 1985.


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% Section 4
\section{EXERCISES.}
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\begin{frame}[allowframebreaks]{D. }

\begin{enumerate}\itemsep2em

\item % Exercise 4.1
Let $M = A_1/A_1x$. Let $\Gamma$ be the filtration of $M$ induced by $\mathcal{B}$ and $\Omega$ the filtration defined by $\Omega_k = B_k \cdot (\partial + A_1x)$. Show that $\text{ann}(M,\Gamma) = S_1y_1$ and $\text{ann}(M,\Omega) = S_1y_1^2 + S_1y_1y_2$. Compute their radicals.

\item % Exercise 4.2
Show that if $J$ is a homogeneous ideal of a graded algebra $R$ then $\text{rad}(J)$ is also homogeneous.

\item % Exercise 4.3
Let $J$ be a left ideal of $A_n$ and put $M = A_n/J$. Show that $\text{Ch}(M) = \mathcal{Z}(\text{gr}(J))$.

\item % Exercise 4.4
Is the union of two {\color{red} involutive} varieties {\color{red} involutive}? What about their intersection?

\item % Exercise 4.5
Show that if $V$ is an {\color{red} involutive} homogeneous variety of $\mathbb{C}^{2n}$ then its {\color{red}irreducible} components are also homogeneous and {\color{red} involutive}.

\item % Exercise 4.6
Let $s$ be the operator of $A_n$ defined in §3. Let $M = A_n/A_ns$.
\begin{enumerate}
    \item Compute $\text{Ch}(M)$.
    \item Is it an {\color{red}irreducible} variety of $\mathbb{C}^{2n}$?
    \item Is it a {\color{red}minimal} hypersurface?
\end{enumerate}

\item % Exercise 4.7
Let $J$ be a left ideal of $A_n$. Show that if $\text{gr}(J)$ is a prime ideal of $S_n$ and $\mathcal{Z}(\text{gr}(J))$ is {\color{red}minimal}, then $J$ is a maximal ideal of $A_n$.

\item % Exercise 4.8
When is a hypersurface of $\mathbb{C}^{2n}$ lagrangian? Give an example of a lagrangian variety of $\mathbb{C}^{2n}$, when $n \geq 2$.

\item % Exercise 4.9
A holonomic left $A_n$-module $M$ is {\color{red} regular} if there exists a filtration $\Gamma$ for $M$ such that $\text{ann}_{S_n} \text{gr}^\Gamma M$ is a radical ideal of $S_n$. Let $N$ be a submodule of a regular holonomic module $M$. Show that $N$ and $M/N$ are also regular.

\item % Exercise 4.10
Show that if $M$ is a regular holonomic module whose characteristic variety is {\color{red}irreducible}, then $M$ is an {\color{red}irreducible} module. Why is the {\color{red} regular} hypothesis required?

\item % Exercise 4.11
Let $\mathcal{J}$ be an ideal of $S_n$. Show that $\mathcal{J}^2$ is always closed for the Poisson bracket.

\end{enumerate}



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